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A Contemporary Anthropological Cosmology!
By!
Ian Beardsley!
Copyright © 2023 by Ian Beardsley"
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Foreward!
One of the things here to point out is that when I have been using the EarthDay in my equation
predicting the unit of a second as natural I have been using the solar day (Earth day with
respect to the Sun) which is a little longer than the sidereal day (Earth day with respect to the
stars). It makes little very difference, but it is actually a little closer to an actual second if you
use the sidereal day. However, I often rounded my equations predicting the unit of a second as
a natural unit both in terms of the atom and the Moon and Earth. So in such instances it makes
no difference whatsoever. In the case of average kinetic energies (Appendix 1) we have!
((3.67E28)/(2.659E22))(86400)=1.1925 seconds!
Which is for the solar day, but for the sidereal day we have!
1.9374 seconds!
For kinetic energies using apogees and perigees (Appendix 1) we have!
(3.428E28)/(2.739E22)(86400=1.08134 seconds!
Which is for the solar day, but for the sidereal day we have!
1.078 seconds!
A sidereal day is 86,164.1 seconds!
The strange thing is that we find the second is like a natural unit for classical and modern
physics in terms of these equations I found for both Moon and Earth and the proton even
though the second has a historical origin that goes back to Ancient times long before these
disciplines existed. Which means we have something interesting to pursue that pertains to
both physics and archaeology. In pursuing the answer to this mystery, it would help to consider
things in terms of both the solar day, and the sidereal day, especially if any work is to be done
from this in developing a method of defining the second in terms of both the proton and the
Earth/Moon/Sun orbital system in terms of not just the calendar that is based on that, but in
terms of the equation I have using the Earth and Moon kinetic energies with the Earth day, and
in its connection to my equation for proton-seconds, which is used to accurately predict the
radius of the proton.!
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Table of Contents!
Abstract……………………………………………………………………4!
1.0 On a Mission to Alpha Centauri………………………………….6
2.0 Alpha Centauri and the Ancestral Adam and Eve……….17
3.0 Earth, Moon, Sun System a Technology………………………..21
4.0 Fertile Crescent And The Moon………………………………….29
5.0 Laying Out The Problem of The Moon As Technology………36
6.0 Constant k and Proton Radius…………………………………45
7.0 The Proton Charge…………………………………………………51
Appendix 1…………………………………………………………..53
Appendix 2………………………………………………………………….54
Appendix 3………………………………………………………………….57
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Cosmology in physics deals with the study of the birth and death of the Universe using the
science of gravity and quanta. But in Anthropology it deals with a People’s way since ancient
times of explaining their relationship to the world around them and the Heaven’s (stars, moon,
sun, and planets). Given today’s data concerning the world and the Universe, I find we can
construct a contemporary cosmology in the anthropological sense. Also..!
Alpha Centauri is the ideal star system for a first interstellar mission because it is the closest
star system to Earth at 4.2 light years. It is a triple system and interestingly two of the stars are
like the Sun Alpha Centauri A and Alpha Centauri B, G and K main sequence stars with masses
1.1 solar masses and 0.9 solar masses respectively. Alpha Centauri C is a small faint red dwarf
but with an Earth sized planet in the habitable zone.!
With my abstract cosmology (Beardsley 2023, Abstract Cosmology and Symmetry Theory) I
predicted with a sixfold basis of reality a great deal including the radius of a proton and the
duration of the life of the Universe in the Standard Friedmann Model by constructing an
equation of state for the periodic table of the elements and a constant k that bridges the
microcosmos with the macrocosmos. Here as an application of the work we suggest that a first
mission to Alpha Centauri may be by design, and that perhaps aliens from, or at least
associated with, this star system were behind the Egyptian pyramids.!
We find we can suggest that the Earth/Moon/Sun system is technology, though we cannot
discern what force would be behind such a design."
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The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
"
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G:6.67408 × 10
−11
N
m
2
kg
2
c : 299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulombs
k
e
= 8.988E 9
Nm
2
C
2
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1.0 On a Mission to Alpha Centauri We suggest the property of matter to have inertia, which
is to say it resists change that the particle a proton, a fundamental unit of matter, is a the cross-
section of a 4D hypersphere in 3D space. As such one could consider inertia as the “friction” of
space given by the normal vector holding the space-bubble in a lower dimensional “plane”:!
Thus the mass of a proton is given by the force of space measured by G, the energy given
to it given by h, the speed at which things happen c, and the surface area of this sphere
that is the cross-section of the hypersphere and the fine structure constant squared
because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 1.1
Thus the equation of inertia (See Appendix 5, Beardsley, Abstract Cosmology and Symmetry
Theory, 2023) can be expressed as proton-seconds. We find that it is six-fold in nature given the
smallest integer for time (1-second) gives carbon the core element of life chemistry, and 6-
seconds is the largest integer before you get fractional protons. That is
Equation 1.2.
Equation 1.3.
m
p
4πr
2
p
α
2
α
2
=
U
e
m
e
c
2
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1pr oton ⋅ 6secon d s = hydr ogen(H )
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$
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting cancel with the bodies of these equations on the left, but
rather divide into them, which are in units of mass, giving us a number of protons. I say this is
the biological because as we shall see our equations are based on one second is 6 protons is
carbon, and 6 seconds is one proton is hydrogen, these making the hydrocarbons which are
the skeletons of biological life. We see this is a mystery of six-fold symmetry based around
biological life in the following computer program I wrote and its output:!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave seconds. We
make a program that looks for close to whole number solutions so we can create a table of
values for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones of biological chemistry. Here is the code for the program:
m
p
1
α
2
m
p
h4π r
2
p
Gc
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#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
We find exactly our equation predicts the second as!
Equation 1.4 !
That this equals so perfectly one second leads us to suggest the second is a Natural Unit. We
find it is, that it is in the kinetic energy of the Moon to that of the Earth times the Earth Day
which is the Earth rotation period:!
Equation 1.5. !
But that was using the average orbital velocities. We find using aphelions and perihelions (See
Appendix 1):!
Equation 1.6.
Interestingly the Moon perfectly eclipses the Sun which means as seen from the earth the Moon
appears to be the same size as the Sun. This is because
Equation 1.7
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352seconds
KE
moon
KE
Earth
(Ear th Day) = 1.2secon ds
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
r
e
r
m
≈
R
⊙
R
m
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Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,
and is the Moon’s radius.. Thus in terms of the Moon and the Sun
Equation 1.8
So we have from
Equation 1.9
Where is the Earth orbital velocity.
Equation 1.10
The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the
Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get to
extremely hot or too extremely cold. We see the Moon may be there for as much of reason as is
Alpha Centauri, for a first manned mission beyond Earth.
Let us now turn to the exact subject of this paper: A Reason For A First Interstellar Mission to
Alpha Centauri. We take equation 1.4
!
And, write it!
Equation 1.11. !
We notice!
!
= !
r
e
r
m
R
⊙
R
m
r
e
=
R
⊙
R
m
r
m
v
e
=
GM
⊙
r
e
v
e
v
e
= G
M
⊙
R
⊙
⋅
R
m
r
m
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352seconds
1
6
⋅
1
α
2
h4π
G ⋅
m
2
p
r
2
p
⋅
m
s
= 1secon d
G
m
2
p
r
2
p
= Ne wtons = Force = (6.67408E − 11)
1.67262E − 27
2
0.833E − 15
2
2.69089E − 34Ne wtons
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!
!
!
Where !
!
!
Putting things in parallel form:!
If , which we will call Universal Force, then!
Equation 1.12. !
And from equation 1.6,…!
Equation 1.13. !
Where is Joule-seconds is like a Planck constant determined by the Earth-Moon system
that is on the macro-scale. .!
We have!
!
!
!
!
We have,…!
h = J ⋅ s = 6.62607E − 34J ⋅ s
c = m /s = 299,792,459m /s
(6.62607E − 34J ⋅ s)4π
(2.69089E − 34N )(299,792459m /s)
= 0.000321273secon ds
N ⋅ m = J
1
α
2
= (137)
2
= 18769
(0.000321273)(18769)
1
6
= 1.004996secon d s
F ⋅
m
2
p
r
2
p
= Force = F
1
6
⋅
1
α
2
h4π
F ⋅ c
= 1secon d
KE
moon
KE
earth
(Ear th Day) =
h
moon
F ⋅ meters
= 1secon d
h
moon
F = 2.69089E − 34N
h
moon
= K E
moon
(Ear th Day)
Ear th Day = (24hrs)(60 min)(60sec) = 86,400secon d s
KE
moon
=
1
2
(7.347673kg)(966m /s)
2
= 3.428E 28J
h
moon
= (3.428E 28J )(86400s) = 2.961792E 33J ⋅ s ≈ 2.96E33J ⋅ s
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!
= !
Let us temper this very large size with smallest size possible using the geometric mean
between it, and the Planck length:!
Equation 1.14. !
Equation 1.15. !
One light year is the distance light travels in one revolution of the Earth and the Sun. Is the
Earth connected to the Alpha Centauri star system? Should we make a real effort to go there?!
!
How many light years is equation 1.15?!
!
Alpha Centauri is 4.2 light years from Earth. We have!
!
If the Earth associates with Alpha Centauri through pi, the ratio of the circumference of a circle
to its diameter ( ), then perhaps there is a reason it is supposed be the first interstellar mission
by way of forces unknown. Interestingly here with the approximations we used, that are
reasonable, we got exactly pi to three places after the decimal. But it is hard to say exactly the
distance to Alpha Centauri because it is a triple system, so where exactly is its center? The
closest of the three stars is Proxima Centauri (Alpha Centauri C) and it is more precisely at
4.2465 light years. But we used 1E67 in equation 1.15, which is more accurately 1.1E67. But
using this value for Proxima Centauri, this compensates for it so it still leaves a good
approximation for pi.!
In that I have!
!
Which is exactly the first three digits of , it is interesting because I luckily rounded values in
doing my computations to get this. You almost always have to round many values. Measured
values can come out a little different depending on the system your ruler uses, because say 1.4
inches may come up on something you measure but with a centimeter ruler measuring the
same length you would have to make an estimate. Some times you always have to round, like
meters =
h
moon
F ⋅ (1secon d )
=
2.96E33J ⋅ s
2.69089E − 34N(1sec)
1.1000E67meters ≈ 1E67meters
ℓ
p
=
ℏG
c
3
= 1.616255(18) × 10
−35
m
(1E67)(1.6E − 35) = 1.265E16meters
1ly = 9.461E15m
1.265E16
9.461E15
= 1.337
4.2ly
1.337ly
= 3.141 = π
π
4.2
1.337
= 3.141
π
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with , because it is irrational and has infinite digits, so to express it as a decimal, even with
each successive decimal being less significant, you have to round at some point. Thus I found
the above ratio interesting because since is irrational you want to approximate it with a
fraction. The above suggests!
and. !
This gives!
!
However the ancient mathematicians according to Maurice Chatelain is his book Our Cosmic
Ancestors, used!
!
He further says they used for the golden ratio!
!
He says with this the Ancient Egyptians reconciled with in the proportions of the Great
pyramid because it gives!
!
That is!
!
And!
!
It was Chatelain who discovered this, though he was not an egyptologist or even archaeologist
but an electronics engineer whose company got the contract to engineer the communications
aspect between Earth and the astronauts for the Apollo missions to the Moon. Since his
clearance gave him insider access he became interested in our knowledge of extraterrestrial
life, and after the missions he wrote his book Our Cosmic Ancestors, where he looked at what
he could formulate about our ancient connections with intelligent life from other worlds. He did
a chapter on the Egyptian pyramids, and said something that pertains to this paper. Really two
things: One, the pyramid had to be 18 million cubits in volume, and why they were obsessed
with this number he didn’t know. In my paper Abstract Cosmology and Symmetry theory I go
into how the basis of Nature is sixfold like we mentioned in this paper: !
π
π
4.2 =
21
5
1.337 ≈
4
3
π ≈
63
20
= 3.15
π ≈
22
7
= 3.142857
Φ ≈
196
121
= 1.619834
π
Φ
4
π
= Φ
1.619834 = 1.272727
4
3.142857
= 1.272727
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And as well I went into 18 as the dynamic factor in sixfold symmetry, like the properties of the
elements are periodic over 18 periods in The Periodic Table of the Elements, because sixfold
comes from the smallest primes 2 and 3, the smallest least common denominators and this gives
And I talk about how this is the dynamic in the world’s music of ancient culture with respect to
meters of 3, 6, and 12. But the other thing he mentioned in Our Cosmic Ancestors since the stars
move due to motions of the Earth that are cyclical the Great Pyramid points at certain sets of
stars at different times, and that 3,400 years BC it pointed to Alpha Draconis (Thuban) which
was the pole star at the time, today it is Polaris but this period of the Earth’s wobble is over
26,000 years. But at this time 3,400 years ago the descending passage in the Great Pyramid
which goes to the Queen’s Chamber, pointed to alpha Draconis but the ascending passage which
goes to the King’s Chamber pointed to Alpha Centauri, the star that is the subject of this paper
as connected to the Earth-Moon system through , just as had to be in the Great Pyramid’s
proportions. One of the other things he suggested was that the pyramids were landing beacons.
That their surfaces were polished and thus reflective, so that the aliens that oversaw their
construction could bounce signals off of them and that, further, the pyramids were flat on top
for landing pads. Whatever the case we are forced to suggest that these aliens came from the
Alpha Centauri System, or are at least associated with it, perhaps using it as base location for
exploration in our region, the nearest star system to us, a triple system with two stars just like
our Sun, and a third red dwarf which we now know has an earth-sized planet in its habitable
zone.
Indeed we can formulate a basis energy for the Universe with the Natural Constants:
Equation 1.16.
And we can formulate a basis velocity, a constant we will call k, by assuming sixfold
symmetry and the orbital velocity of the Earth:
Equation 1.17.
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1pr oton ⋅ 6secon d s = hydr ogen(H )
2 ⋅ 3 = 6
3
2
= 9
2 ⋅ 9 = 18
3 ⋅ 6 = 18
π
π
h
G
⋅
c
3
m
p
= 1.599298E 29J
k v
e
= 6
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I have derived this constant k in another paper and used it to determine the radius of a proton
and its charge (Abstract Cosmology And Symmetry Theory, Beardsley 2023):
Equation 1.18.
Where is Avogadro’s number is 6E23 atoms per gram and is 1 gram per atom when
hydrogen. I have come to find this value of k is closest to :
Equation 1.19.
We want to use the form of our equation
In the next step to bridge the microcosmos to the macrocosmos we create an intermediary mass
midway between the proton mass and the upper limit of mass for a white dwarf star to
form without collapsing into a blackhole star. It balances with its gravity by radiation pressure
alone, the so-called Chandrasekhar limit:
Equation 1.20.
We make the approximation and define with the geometric mean:
Equation 1.21
Giving
Equation 1.22.
Using , we have
Equation 1.23.
Because in my paper (Abstract Cosmology And Symmetry Theory, Beardsley 2023)
Equation 1.24.
We had
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
N
A
𝔼
1/k = 788.46m /s
k v
e
=
29790m /s
788.46m /s
= 6.145748
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
m
i
m
p
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
0.77 ≈ 3/4
m
i
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
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We can hone this by reintroducing 0.77 for 3/4
Thus precisely:
We have
We have honing our
Equation 1.25.
Now we use our equation
Which has kinetic energy over kinetic energy we have
Equation 1.26.
This is
Equation 1.27.
Equation 1.28.
1 Earth year = (365.25)(24)(60)(60)=31557600 seconds
The universe is theorized by standard models to die in 100 trillion years, which is when the last
stars born will die out. This is exactly 1E14 yearsWe see the moon is connected to how old the
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
K E = m
i
(
1
k
)
2
K E = (68.897kg)(788.4626m /s)
2
= 42,831,358Joules
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) =
1.599298E 29J
42831358J
= 3.734E 21secon ds
3.734E 21s
31557600s
= 1.1832332E14years ≈ 1E14years
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universe is theorized to become. Indeed there may be an infinite number of Universe being born
and reborn at different times. Consider we started out as a small single cell organism and
became advanced intelligence in 4 billion years. Before the Universe ends in 100 trillion years I
would think we wouldn’t have physical bodies and could travel across any distance with our
mind to newly born Universes, even parallel universes in other dimensions. But what we have is
Equation 1.29.
Equation 1.30.
We suggest for some mass , we have
Equation 1.31.
It is given by
Equation 1.32.
For all practical purposes this is the mass of the Moon, which is exactly 7.34767E22kg.
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
M
h
G
⋅
c
3
m
p
M
(
1
k
)
2
(1secon d ) = 1Ear thYear
M =
h
G
⋅
c
3
m
p
(1Ear thYear)
(
1
k
)
2
(1secon d )
h
G
⋅
c
3
m
p
=
6.62607E − 27
6.67408E − 11
⋅
(299792459)
3
1.67262E − 27
= 1.599E 36J ≈ 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
1Ear thYear = (365.25)(24)(60)(60) = 31557600s
M =
1.599E 36
621673
⋅
1
31557600
= 8.15E 22kg
8.15E 22kg
7/34767E 22kg
= 1.10919516 ≈ 1.12m oon s
of 17 60
Equation 1.33.
2.0 Alpha Centauri and the Ancestral Adam and Eve
The hypothesis in my paper would naturally have as the next step a look at where Alpha
Centauri is in the Earth sky. Well it is in the Southern Hemisphere and visible highest in the sky
from the Southernmost tip of South Africa, from the Southernmost tip of Argentina, and from
New Zeland and Australia. The Southernmost tip of Africa is where we have traced the origin of
anatomically modern humans. The so-called Ancestral Adam and Eve. Continue to next page…
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
of 18 60
Alpha Centauri ( Centauri) is at declination meaning it passes directly overhead at
latitude which is right through the Drake Passage through which one must pass by ship going
from the Southernmost point of South America to the to the Antarctica landmass where the
South Pole is. One takes a ship from Argentina, the closest part of the landmass to the
Antarctica, and crosses the Drake Passage to the South Shetland Islands and from their one
can get to the main South Antarctica Landmass which is 98% covered in ice, has no villages,
towns, or cities—Just scientific research stations.!
Alpha Centauri then, as the Earth rotates, traces out a path around the outer periphery of the
Antarctica Mainland. The South Shetland Islands, which are considered part of Antarctica, were
discovered in 1819 by the British Mariner William Smith and the islands were claimed by
England but by the Antarctic Treaty of 1959, the islands are free for use by any signatory for
non-military uses. The island were named by a Russian explorer after the islands that make the
Northernmost territory of Scotland, the Shetland Islands, which are just as far north of the
equator at latitude as their counterpart in the Antarctica is South. In contrast these
islands of Scotland, which were once part of Norway, but given to Scotland in a royal dowry
dispute, are very inhabited, with may villages and thousands of Ancient Mesolithic sites, and a
rich cultural history, and are much warmer than their Antarctic counterpart because the Arctic
is warmer than the Antarctic because in Antarctic the winds are much stronger.!
Drake Passage with
60th Parallel South
passing through it.!
α
−61
∘
−61
∘
+60
∘
of 19 60
Alpha Centauri is at!
RA: !
Dec: !
For all practical purposes at South which passes through the Drake Passage between
Argentina and the Antarctica land mass near the South Shetland Islands and its journey traces
out a line forming the periphery of the Antarctic. If we go North, this star will get lower in the
sky until we reach the first continent which is Africa and the first place we reach is South Africa
around Botswana. This is at the 30th parallel South, latitude meaning we have travelled
60/2=30 degrees North correspondingly meaning Alpha Centauri has dropped 30 degrees in
the sky from the zenith putting it at putting it 90-30 equals 60 degrees above
the horizon in Botswana, a parallel that passes through Chile, Argentina, and Australia.!
Botswana is where genetic anthropology has put the earliest anatomically modern humans
from which we all descended. They are called the Ancestral Adam and Eve. Fortunately we can
know something about the astronomy that Ancient Africans had because they passed down
some of their formulation of what the Sun, Moon, and stars meant to them, what the
relationship is they had with the heavens. Indeed they knew women menstruated every 28 days
just as the Moon cycled through its phases with about the same period. That the Moon is a
natural clock for this is interesting in itself. Alpha Centauri is the third brightest star in the sky,
and the Earth is the third planet from the Sun. But, so bright it, and Centauri, the second
brightest star in the constellation (Alpha Centauri is the first) it played a prominent role in their
cosmology. Other important stars are those of the Southern Cross where and Centauri are
its pointers. By the Sotho, Tswana, and Venda peoples, who are South African peoples in or
around the Botswana region where our 30th parallel South is, these stars are called the
14h39m37s
−60
∘
50′ 2′ ′
−61
∘
−30
∘
60 − 30 = 30
∘
β
α
β
of 20 60
Dithutlwa, meaning “The Giraffes”. The brightest stars of the cross are the male giraffes and
the two pointers are the female giraffes. The Venda say the month Khubuhumezi, the month
where the cultivating season begins, is when the lower two giraffe stars are just below the
horizon, and upper two are just above it, and the Moon is a crescent for the first time."
of 21 60
3.0 Earth, Moon, Sun System a Technology
As further development in this paper I find it worth suggesting that the Earth/Moon/Sun system
is a technology geared towards life though it is not discernible as to what force would be
behind it."
of 22 60
In so far as we have!
Equation 3.1. !
It is to say that the kinetic energy of the Moon to the kinetic energy of the Earth is equal to the
radius of a proton to its mass times a factor k!
Equation 3.2. !
We have!
!
!
!
So its units are!
!
Giving a value of!
!
!
And verifying that it works!
!
!
KE
moon
KE
earth
(Ear th Day) =
r
p
m
p
⋅
1
6α
2
h4π
Gc
k =
1
6α
2
h4π
Gc
h = J ⋅ s =
(
kg
m
s
2
⋅ m ⋅ s
)
=
kg ⋅ m
2
s
G = N
m
2
kg
2
= kg
m
s
2
⋅
m
2
kg
2
=
m
3
s
2
⋅ kg
c =
m
s
h
Gc
=
(
kg ⋅ m
2
s
)(
s
2
⋅ kg
m
3
)
(
s
m
)
=
kg
2
⋅ s
2
m
2
=
kg ⋅ s
m
1
6α
2
h4π
Gc
=
(6.62607E − 34)(4π)
(6.67408E − 11)(299,792,459)
⋅ 3128.167 = 2.0179798E − 12
≈ 2E − 12
kg ⋅ s
m
r
p
m
p
=
0.833E − 15m
1.67262E − 27kg
4.9802E11
m
kg
r
p
m
p
k = (4.9802E11)(2E − 12) = 0.99604secon ds ≈ 1secon d
of 23 60
Since we are dealing with some sort of a link between the microcosmos and the macrocosmos
we want to form some kind of a deBroglie wave length associated with the mass of a proton:!
deBroglie wavelength: !
Equation 3.3 !
!
The radius of the sun is meters giving it a diameter of .
We take the ratio between this and our deBroglie wavelength to get!
!
But the radius of the Sun is difficult to pinpoint. Beyond the bulk of its mass it has a thin
atmosphere that reaches far out into space. However, the Moon nearly perfectly eclipses the
Sun as seen from the Earth allowing us to study its outer atmosphere during an eclipse. The
moon nearly perfectly eclipses the Sun as seen from the Earth because the Sun may be 400
times larger than the Moon, but the Moon is 400 times further from the Sun than it is from the
Earth. This is to say that!
Equation 3.4. !
Where is the Earth orbital radius, is the lunar orbital radius, is the solar radius, and
is the lunar radius. So why not let the size of the Moon as seen from the Earth define the radius
of the Sun by writing!
Equation 3.5. !
The Moon at closest approach to Earth (perigee) is 1.7371E7 meters:!
!
This gives!
!
We have!
λ =
h
mv
k
h
m
2
p
=
(
kg ⋅ s
m
)
(
kg ⋅ m
2
s
)
(
1
kg
2
)
= λ = meters
(2.018E − 12)(6.62607E − 34)
(1.67262E − 27)
2
= 4.7795E8m
R
⊙
= 6.95700E8m
D
⊙
= 1.3914E 9m
1.3914E 9
4.7795E8
= 2.911 ≈ 3
r
e
r
m
≈
R
⊙
R
m
r
e
r
m
R
⊙
R
m
R
⊙
=
r
e
r
m
R
m
1.496E11m
3.632289E8m
⋅ 1.7371E6m = 7.154446E8m
D
⊙
= 1.430889E 9m
of 24 60
!
We want three because it is the ratio of the perimeter of a regular hexagon to its radius. In a
regular hexagon its sides equal its radii, meaning with a perimeter of 6 (It is six-sided) its
diameter is 2, and 6/2=3. This is the recurrence of our sixfold symmetry. Three is the hexagonal
approximation to .!
Thus we have!
Equation 3.6. !
Equation 3.7. !
Equation 3.8. !
Equation 3.9. !
!
Equation 3.10. !
Equation 3.11. !
We know that!
!
So we have!
Equation 3.12. !
Or!
1.430889E 9
4.7795E8
= 2.9938 ≈ 3
π
2R
⊙
k
⋅
m
2
p
h
= 3
2R
⊙
= 2
r
e
r
m
R
m
2
r
e
r
m
R
m
⋅
m
2
p
hk
= 3
r
e
r
m
R
m
m
2
p
hk
=
3
2
k =
1
6α
2
h4π
Gc
r
e
r
m
R
m
m
2
p
h
6α
2
Gc
4πh
=
3
2
m
2
p
r
e
r
m
R
m
α
4
π
Gc
h
3
=
1
2
R
⊙
=
r
e
r
m
R
m
m
2
p
R
⊙
α
4
π
Gc
h
3
=
1
2
of 25 60
Equation 3.13. !
!
!
So we know equation 13 is indeed unity on the right. Remember since the sun has a thin
atmosphere that goes far out into space we are defining its radius in terms of the Moon, and
hence its diameter so!
Equation 3.14. !
Let us see how accurate the equation is.!
!
!
= !
!
Which is near 100% accuracy. So, essentially, our system of equations!
!
Which define the second as a natural unit, and give!
!
Say that the diameter of the Sun is inversely proportional to the square of the mass of a proton:!
m
2
p
D
⊙
α
4
π
Gc
h
3
= 1
Gc
h
3
=
(
m
3
s
2
⋅ kg
)
(
m
s
)
(
s
3
kg
3
⋅ m
6
)
=
1
m ⋅ kg
2
m
2
p
D
⊙
= kg
2
m
D
⊙
= 2
r
e
r
m
R
m
m
2
p
D
⊙
= (1.67262E − 27)
2
(1.430889E 9) = 4.0031E − 45
α
4
π
Gc
h
3
=
2.838688E − 9
3.141592654
⋅
(6.67408E − 11)(299,792,459)
(6.62607E − 34)
3
2.4929E44
(4.0031E − 45)(2.4929E44) = 0.9979 ≈ 1
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352seconds
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
KE
moon
KE
earth
(Ear th Day) =
r
p
m
p
⋅
1
6α
2
h4π
Gc
of 26 60
!
Where G, c, h, , and are the properties of space, time, and matter that make this possible. If!
Equation 3.15. !
Then!
Equation 3.16. !
The kinetic energy of the Earth in its orbit around the Sun is due to the mass of the Sun and its
distance from it. The kinetic energy of the Moon is due to the mass of the Earth and its
distance from it. The Moon makes life on Earth possible because its orbit around the Earth
holds the Earth at its tilt to its orbit allowing for the seasons and preventing extreme
temperature of hot or cold. In light of what we have found I would guess the Earth/Moon/Sun
system is a technology that allows for life. What force might be behind such a technology is
indiscernible at this point.!
Instead of letting k represent the expressions in equations 3.2 and 3.15, let us use and :!
!
!
We write Equations 3.2 and 3.16 as!
!
!
Which yield!
!
We have!
m
2
p
D
⊙
α
4
π
Gc
h
3
= 1
α
π
k =
α
4
π
Gc
h
3
k D
⊙
=
1
m
2
p
A
B
A =
1
6α
2
h4π
Gc
B =
α
4
π
Gc
h
3
KE
moon
KE
earth
(Ear th Day) = A
r
p
m
p
BD
⊙
=
1
m
2
p
D
⊙
=
KE
2
moon
KE
2
earth
(Ear th Day)
2
1
A
2
B
⋅
1
r
2
p
of 27 60
!
!
And, we say that!
!
Yielding!
Equation 3.17. !
Equation 3.18. !
3.18 has units!
!
3.17 has units!
!
!
We just need to compute our values:!
!
!
= !
Computations of Kinetic energies of the Moon and Earth are in Appendix 1:!
A
2
=
1
36α
4
⋅
h4π
Gc
A
2
B =
1
9α
2
π
Gch
C = A
2
B
D
⊙
=
KE
2
moon
KE
2
earth
(Ear th Day)
2
1
Cr
2
p
C =
1
9α
2
π
Gch
π
Gch
=
(
s
2
⋅ kg
m
3
)
(
s
m
)(
s
kg ⋅ m
2
)
=
s
2
m
3
KE
2
moon
KE
2
earth
(Ear th Day)
2
r
2
p
=
s
2
m
2
s
2
m
2
⋅
m
3
s
2
= m eters
C =
1
9α
2
π
Gch
=
18769
9
3.141592654
(6.67408E − 11)(299792459)(6.62607E − 34)
1.015E 21
of 28 60
We had in terms of the Moon eclipsing the Sun
Which is 86% accuracy."
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
Ear th Da y = (24)(60)(60) = 86400secon ds
r
2
p
= (0.833E − 15m)
2
= 6.93889E − 31
K E
2
moon
K E
2
earth
(Ear th Da y)
2
r
2
p
=
3.428E 28
2
2.7396E 33
2
⋅
86400
2
6.93889E − 31
= 1.6844E 30
1.6844E 30
1.015E 21
= 1.6595E 9m eters
D
⊙
= 1.430889E 9m
of 29 60
4.0 Fertile Crescent And The Moon!
of 30 60
We said that the Earth rotation axis is tilted 23.5 degrees to its orbital plane around the Sun,
that is, to the ecliptic, the path of the Sun due to the Earth’s annual motion around the Sun.
And we said the reason the Earth stays tilted to its orbit is that the Moon’s orbit around the
Earth holds the tilt in place. We said this creates the seasons and allows for life on Earth to be
abundant, diverse, and stable because it prevents extreme hot or cold. We would like to go into
this in depth because in the previous section we found reason to suggest the Earth/Moon/Sun
system is technology that may be geared towards the success of life.!
If the earth is tilted towards the Sun by 23.5 degrees this is summer and if it is tilted away by
23.5 degrees this is winter. At fall and spring equinox the Earth is tilted neither towards or away
from the Earth. However, the seasons are more extreme as you move North of the equator.
Because the equator is only 23.5 degrees above or below the ecliptic, the plane of its orbit, it is
always facing the Sun so within this region, 23.5 degrees North of the equator and 23.5
degrees South of the equator, the Sun is always high in the sky at noon, meaning there is little
difference between summer and winter. Even the days are always approximately equal in
duration to the nights. They are both 12 hours long. This is a region called the Tropics. As you
travel North or South of the Tropics the seasons become more notable. When the Earth is
inclined towards the Sun it is summer in the Northern hemisphere and winter in the Southern
hemisphere where it is pointed away from the Sun. When the Earth moves around to the other
side of the Sun, the Earth is tilted away from the Sun in the Northern hemisphere but is tilted
towards the Sun in the Southern hemisphere and it is summer. Depending on your latitude
above or below the equator and the time of the year (the Earth’s position with respect to the
Sun in it its orbit) determines how high the Sun is in the sky during the day and thus amount of
sunlight you get and the season. The time of year determines in these latitudes the hours in a
day and the hours in a night. So at higher latitudes North or South of the tropics means in the
summer you can have 16 hour days with 8 hour nights in the Northern latitudes. Or, in the
winter 8 hour days with 16 hour nights. In fact in the Arctic and Antarctic it is daylight for 6
months and night for six months.!
We can determine the intensity of sunlight given the time of the year and your latitude. On the
summer solstice, the Sun is directly overhead at noon 23.5 degrees above the equator (The
Tropic of Cancer, or Northern Tropic). Your latitude minus this angle is your solar angle, the
degrees the Sun is below the Zenith (see illustration next page) or the height it is in your sky at
noon. The cosine of this angle reduces the intensity of the Sun giving its value for your location
at this time. On winter solstice your solar angle is your latitude plus 23.5 because the Sun is
now overhead at noon 23.5 degrees below the equator (The Tropic of Capricorn, or Southern
Tropic) and the intensity of the Sun is reduced by the cosine of this angle.!
For our purposes, since we are suggesting The Earth/Moon/Sun system is technology, and that
the Moon holds the Earth at 23.5 degrees to the ecliptic, then if this angle is significant as part
of the technology, we want to ask: At what latitude is agriculture (which provides sustenance
for life) optimized? Of course we have different soils at different latitudes, different amounts of
sunlight and seasonal variations, but history answers the question, actually archaeology. We
ask at what point did humans settle down from following the herds and hunting with stone
spearpoint, and invent agriculture, mathematics, smelting, government, and writing. The
answer is in Mesopotamia (Sumer) twelve to fourteen thousand years ago. This was part of the
fertile crescent which spans from Egypt to Turkey where there are many archaeological sites
showing the beginning of agriculture. Scholars put the fertile crescent at latitude 33.7 degrees
though it can be as far South as Egypt or as far North as Turkey. We have for the solar angle
and intensity of sunlight at this latitude:!
!
!
Summer : cos(33.7 − 23.5) = 0.9842
Winter : cos(33.7 + 23.5) = 0.5417
of 31 60
"
of 32 60
!
!
Thus in the summer the fertile crescent gets 98.42% of the sunlight that reaches it and in the
winter 54.17%. This is a ratio of!
!
Which is about 1.8 times more sunlight in the summer than in the winter. I recognize this ratio.
It is the Solar radius to the Lunar orbital radius which is the ratio of an atom of gold to an atom
of silver by mass, and the Sun is gold in color and the Moon is silver in color. Gold and silver
have been at the heart of ceremonial jewelry since ancient times.!
We want to consider how much sunlight reaches the Earth. The solar luminosity is:!
!
The separation between the Earth and the Sun is!
!
The solar luminosity is reduced at Earth by the inverse square law:!
!
Thus in the summer, the fertile crescent receives!
!
And in the winter:!
!
intercepts the Earth disc and distributes itself over the entire surface and
because the Earth’s albedo, , is 0.3 it reflects 30% of the light back into space. We have!
!
Temperature to the fourth is proportional to radiation by the Steffan-Boltzmann constant, :
means temperature entering:!
!
Summer : cos(33.7 − 23.5) = 0.9842
Winter : cos(33.7 + 23.5) = 0.5417
98.42
54.17
= 1.81687
L
0
= 3.9E 26J/s
r
e
= 1.5E11m
S
0
=
3.9E 26
4π(1.5E11m)
2
= 1370
Wat ts
meter
2
(0.9842)1370 = 1348.354
Wat ts
meter
2
(0.5417)1370 = 742.129
Wat ts
meter
2
S
0
πR
2
4πR
2
a
(1 − a)S
0
πR
2
4πR
2
= (1 − a)
S
0
4
σ
T
e
σ T
4
e
=
S
0
4
(1 − a)
of 33 60
So we have the annual average temperature would be!
!
!
!
The solar constant is usually 1361 (Well established) because we rounded our value for the
Earth distance and the Earth orbit is not perfectly circular. Let’s convert this to centigrade:!
!
!
The actual annual average temperature of the Earth is 14C or 57F. But we have not included
the effect of the Earth’s greenhouse gases holding in heat. In a one layer atmosphere model we
consider that The radiation entering the system equals the radiation leaving the system when
the Earth is in radiative equilibrium. The atmosphere radiates back to the surface. So
the temperature at the surface of the Earth is!
!
So,…!
!
!
!
!
So this is hotter than the annual average temperature we said was 14C or 57F but we have not
considered cooling by convection; A lot of the radiation goes into warming the ocean, which
has a high specific heat (4.184J/g-K) and when this water evaporates it precipitates and returns
as rain. If we include this we get much closer to accurately predicting the global temperature.
There are other cooling mechanisms to consider as well. The average temperatures in Turkey (a
representative of the fertile crescent) is!
Summer: 26-29 C, 79-84 F!
Winter: 7-10 C, 45-50 F!
That is annual average temperatures of 17-20 C, 63-68 F!
T
e
=
4
S
0
4σ
(1 − a)
σ = 5.67E − 8
T
e
=
4
1361
4(5.67E − 8)
(0.7) = 254.58K
T
e
= 254.58K − 273.15K = − 18.57
∘
C
(−18.57)(1.8) + 32 = − 1.42
∘
F
T
a
= T
e
σ T
4
S
= σ T
4
e
+ σ T
4
a
= 2σ T
4
e
T
S
= 2
1
4
T
e
T
S
= (1.18921)254.58 = 302.75K
T
S
= 302.75 − 273.15 = 29.6
∘
C
29.6(1.8) + 32 = 85.28
∘
F
of 34 60
The annual average temperature globally we said was 14C or 57 F, so Turkey on the annual
average is hotter than the average globally by about 4 C or 9 F. The Global Annual Average
average, is the average temperature annually from points all over the globe, everywhere thus
including the very cold arctic regions.!
That we computed the annual average temperature to be!
!
!
!
And say with cooling by convection (warming of the ocean) we get close to our 14C or 57 F is
based on the idea the Earth is in radiative equilibrium (loses as much as it gains). If the Earth
gains more than it loses it is warming, if it loses more than it gains it is cooling. The Earth is
currently gaining about one watt per square meter more than it loses which means it is
warming. The chief culprit is the greenhouse gas CO2. Trees create CO2 and it is needed for
the Earth to not freeze. We said the annual average temperature without the atmosphere
radiating heat back to the surface would be!
!
!
!
Not all gases are greenhouse gases, like the primary constituents of the atmosphere like
nitrogen gas N2 and oxygen gas O2. But methane is a greenhouse gas that holds more heat
than CO2, but the warming is mainly due to CO2 because there is a lot more of it. What we
mean by green house gases are gases that radiate heat absorbed back to the Earth’s surface.
And indeed we need the CO2 created by plants and trees to keep the Earth warm enough, but
too much and the Earth is out of equilibrium meaning the temperature is constantly rising
taking us beyond what the annual average should be. The plants and trees absorb some of the
CO2 and make from it the oxygen that we breathe. But we are putting beyond what it should
be in the atmosphere by burning fossil fuels for transportation and industry, and coal to make
electricity. It is computed the tipping point for the Earth to go into catastrophic warming is 350
parts per billion CO2 in the atmosphere. We are currently at about 400. The solution is to
change to alternate forms of energy like wind and solar. We have many technologies that could
resolve this problem, which would actually meet our energy needs at a cheaper price, but we
have failed to implement them for political reasons.!
We said the earth receives the most sunlight of the year on the summer solstice around June
20 or June 21, when the Earth is tilted most towards the Sun the most, when the sun is directly
overhead at noon at the latitude 23.5 degrees, the Tropic of Cancer. However this is not the
hottest day of the year because it takes time for the Earth to warm and for it to cool, so there is
a lag and the hottest time comes in the months following the solstice around August and
T
S
= (1.18921)254.58 = 302.75K
T
S
= 302.75 − 273.15 = 29.6
∘
C
29.6(1.8) + 32 = 85.28
∘
F
T
e
=
4
1361
4(5.67E − 8)
(0.7) = 254.58K
T
e
= 254.58K − 273.15K = − 18.57
∘
C
(−18.57)(1.8) + 32 = − 1.42
∘
F
of 35 60
September. The temperature is determined by the heat going in minus the heat going out, and
the oceans don’t start too cool enough until after August and September when a lot of heating
is still going in.!
The principle climatic force is the oceans. It stores radiation from the Sun. The oceans store
heat from the Sun, mostly in the Tropics where there is more ocean and the intensity of the Sun
is highest. Water here is constantly evaporating to form rain and storms that are carried by
trade winds to the land. Weather patterns are driven by ocean currents created by surface
winds and the rotation of the Earth. Current system flow clockwise in the northern hemisphere
and counterclockwise in the southern hemisphere. These ocean currents distributing the effect
of solar radiation more evenly over the planet transporting warm weather and precipitation to
the poles and cold water from the poles to the tropics. The currents prevent extreme heat at
the equator and extreme cold at the poles.!
Convection is a key science to understand in global warming science, it falls under a physics
course in thermal dynamics. Convection quite simply is the motion of gases and liquids such
as the atmosphere and water. Temperature is a measure of the motion of atoms. That is heat is
the excited atoms bouncing off one another. A warmer substances transfers heat to
surrounding cooler substances, an effect known as heat transfer. It does this because the
warmer atoms are moving more and thus collide with the atoms of the cooler surrounding
substances giving them more motion, or increasing the temperature of these substances, but
from these collisions losing motion, and thus cooling themselves until the entire medium goes
into equilibrium, and evenly distributes temperature that is cooler than what the warmer region
was to begin with.!
More than 93% of the excess heat energy from climate change is absorbed by the oceans
according to the IPCC. In 2022 the oceans absorbed 9 zetajoules of heat from the atmosphere.
The ocean heat content (OHC) is given by a definite integral that sums over the heat at different
layers or depths of the ocean. That is the heat, H, is given by summing over all of the layers:!
!
Where is the specific heat capacity, the amount of energy in Joules required to raise a gram
one degree centigrade of sea water. And, h2 is the lower depth, h1 the upper depth, is the
seawater density profile, and is the temperature profile. H has the units of Joules per
square meter.!
H = c
p
∫
h1
h2
ρ(z)T (z)d z
c
p
ρ(z)
T(z)
of 36 60
5.0 Laying Out The Problem of The Moon As Technology
In laying out the problem of the Moon as technology we consider what we have found. Namely!
Equation 5.1. !
Which we showed has nearly 100% accuracy. Where the solar diameter is given by the orbital
radius of the earth , the orbital radius of the Moon and the radius of the moon :!
Equation 5.2. !
We can say!
Equation 5.3. !
So we can write:!
Equation 5.4. !
Which means the diameter of the Sun is inversely proportional to the square of the mass of a
proton, the primary particles of the Sun are protons and neutrons which have about the same
mass. We also said!
Equation 5.5. !
Where!
Equation 5.6. !
We found the equation has 86% accuracy. We list this equation because it is the kinetic energy
of the Moon that holds the Earth in its orbit with its tilt of 23.5 degrees that allows for the
success of life. We concluded that this tilt puts agriculture, the fertile crescent at the latitude of
33.7 degrees North that optimizes it. We have for intensity of sunlight with latitude (lat):!
Equation 5.7. Winter: !
Equation 5.8. Summer: !
Where is the solar intensity and is its intensity at the latitude in question. Thus we want to
look at the angle of 23.5 degrees, the tilt of the Earth provided by the Moon which seems to
optimize conditions for the success of life:!
m
2
p
D
⊙
α
4
π
Gc
h
3
= 1
r
e
r
m
R
m
D
⊙
= 2
r
e
r
m
R
m
k =
α
4
π
Gc
h
3
k D
⊙
=
1
m
2
p
D
⊙
=
KE
2
moon
KE
2
earth
(Ear th Day)
2
1
Cr
2
p
C =
1
9α
2
π
Gch
cos(lat + 23.5
∘
)I
0
= I
s
cos(lat − 23.5
∘
)I
0
= I
s
I
0
I
s
of 37 60
Equation 5.9. !
Equation 5.11. !
Where 90 degrees is the maximum sunlight through a surface and 45 degrees splits it evenly.!
Equation 5.12. !
We have said equation 1.1 !
!
is the diameter of the Sun which specifies its size but the convention we specify its size
with is its radius . You can do it either way since it is taken as spherical. The luminosity of
the Sun is given by its size because the greater the radius the surface area it has to radiate
light. The luminosity of the Sun, or any star, is given by!
Equation 5.13. !
Luminosity, L, is in joules/second which is determines our solar constant .
And, R is and the is because luminosity is temperature to the fourth by the Stefan-
Boltzman constant . The Sun has a temperature of and the
radius of the Sun is 6.95700E8m is approximately is 6.96E8m. Thus!
!
So we see this works pretty well. The temperature of a star is related to its mass (see Appendix
3):!
Equation 5.14. !
So from equation 5.13:!
Equation 5.15. !
From equation 5.4!
23.5 ≈ 45/2 = 22.5
45 = 90/2
cos(45
∘
) =
2
2
= sin(45
∘
) =
2
2
m
2
p
D
⊙
α
4
π
Gc
h
3
= 1
D
⊙
R
⊙
L = 4πR
2
σ T
4
S
0
= 1361W/m
2
R
⊙
T
4
σ = 5.67E − 8J/(m
2
⋅ K
∘
)
5700
∘
K
L = 4π(6.96E8m)
2
(5.67E − 8)(5700
∘
K )
4
= 3.64E 26J/s = 3.64E 26W
T =
(
e
2
4πϵ
0
)
2
m
p
3π
2
h
2
k
B
L
⊙
= 4πR
2
⊙
σ
(
e
2
4πϵ
0
)
8
(
m
p
3π
2
h
2
k
B
)
4
of 38 60
!
Or,…!
!
So equation 5.15 becomes!
Equation 5.16. !
And, we have described the luminosity of the Sun in terms of our equation 5.1 which is!
!
!
This is what we want to do.!
Thus, in the fertile crescent the Moon has a change in radiation over a year because the Moon
holds the Earth at its tilt of 23.5 degrees, of!
!
Which results in a temperature change over a year of!
!
Which is a change equal to room temperature.!
The pressure at which water passes from its liquid phase to it gaseous phase (vapor) is called
vapor pressure. There are three states for water solid, liquid, and gas. These states, or phases,
mostly depend on temperature and pressure.!
If air that contains moisture is heated, the pressure of the moisture increases exerting pressure,
and when this increases to a point when when the number of water molecules evaporating
equals the number condensing (returning to liquid after cooling) we say a point of saturation
has been reached. This point defines the boiling point of water which is 100 degrees
centigrade. The saturation temperature is related to a change in pressure that has
brought about a change in temperature . The work done is!
k D
⊙
=
1
m
2
p
R
⊙
=
1
2k m
2
p
L
⊙
= π
(
1
k m
2
p
)
2
σ
(
e
2
4πϵ
0
)
8
(
m
p
3π
2
h
2
k
B
)
4
m
2
p
D
⊙
α
4
π
Gc
h
3
= 1
k =
α
4
π
Gc
h
3
(0.9842)(1361W/m
2
) − (0.5417)(1361W/m
2
) = 1339.5 − 737.25 = 602.25W/m
2
= ΔS
ΔT = 47.5
∘
− 27.5
∘
= 20
∘
= 68
∘
F
T
ΔP
ΔT
of 39 60
Equation 5.17. !
Where C is the latent heat of water. And for the change in state from liquid to gas has resulted
in a change in volume given by the change in pressure :!
Equation 5.18. !
This means we have!
Equation 5.19. !
Which is the differential equation!
Equation 5.20. !
This last equation is called the Clausius-Clapeyron relationship. A solution for this equation if
you are not working with the transition from solid to liquid, is!
Equation 5.20. !
Which is upon integration!
Equation 5.21. !
Where P1 and P2 are the vapor pressures at T1 and T2 and R is the gas constant
. And is the energy of vaporization which is which
can vary a little with temperature, but this works as a good approximation. These equation are
pivotal to meteorology (predicting weather, like rain). The reason this predicts rain is that while
sunny skies are associated with high pressure, rain is associated with low pressure. The reason
is that high pressure holds down the moisture, but if there is low pressure the the air with
moisture rises and forms clouds. Thus you know it is going to rain if your barometer reads a
sudden drop in pressure.!
The atmospheric pressure is caused by the force of the atmosphere upon a square area.
Therefore it is a function of the density of air as acted on by the gravitational accleration g. The
density decreases with altitude as well as the amount of air piled up on the that altitude, so it is
a function of h, the height above sea level. It is also a function of temperature because that
affects the pressure of the atmosphere. The density can be described by the molar mass of air
which is basically 21% O2 and 78 percent N2. At low altitudes above sea level the atmosphere
is 1.2 kPa for every 100 meters. The barometric formula gives the atmospheric pressure as a
function of h. It is!
ΔW = C
ΔT
T
ΔP
W = ΔP(V
g
− V
l
)
ΔP
ΔT
=
C
TΔV
∂P
∂T
=
C
TΔV
P ∝ exp
(
−
ΔH
vap
RT
)
ln
(
P
1
P
2
)
=
ΔH
vap
R
(
1
T
1
−
1
T
2
)
8.3145J ⋅ mol
−1
⋅ K
−1
ΔH
vap
40.7k J/m ol
of 40 60
!
= !
!
Where h = height above mean sea level=m, !
sea level standard atmospheric pressure=101,325Pa!
L=temperature lapse rate= for dry air is~0.00976 K/m!
constant-pressure specific heat=1,004.68506 !
Sea Level Standard Temperature 288.16K!
Earth-surface gravitational acceleration= !
molar mass of dry air=0.02896968kg/mol!
Universal Gas Constant=8.314462618 !
A barometer reading of 30 inches is considered normal, and strong high pressure could
measure as high as 30.70 inches. Low pressure, like that associated with a hurricane can be
below 27.30 inches. Hurricane Andrew measured 27.23 just before hitting land in Miami Dade
County. Generally a reading below 29.80 is considered low. We can use!
!
To compute the atmospheric pressure at sea level by letting where is the radius
of the Earth and is the mass of the atmosphere. . And
. This gives!
!
!
!
p = p
0
(
1 −
L ⋅ h
T
0
)
g ⋅ M
R
0
L
p
0
(
1 −
g ⋅ h
c
p
⋅ T
0
)
c
p
⋅ M
R
0
≈
p
0
⋅ exp
(
−
g ⋅ h ⋅ M
T
0
⋅ R
0
)
p
0
=
g/c
p
c
p
=
J/(kg ⋅ K )
T
0
=
g =
9.80665m /s
2
M =
R
0
=
J/(m ol ⋅ K )
PA = mg
A = 4πR
2
⊙
R
e
m = M
atm
M
atm
= 5.1480E18kg
g = 9.8m /s
2
P =
M
atm
⋅ g
4πR
2
e
R
e
= 6378.1k m = 6,378,100
P =
(5.1480E18)(9.8)
4π(6378100)
2
= 98689.762N/m
2
of 41 60
The atmospheric pressure is given in the data tables as So our estimate is very
good:!
!
Is an accuracy of 97%. But, for our purposes we want this in terms of the mass and radius of
the Earth using!
!
Equation 5.23. !
Now that we have an equation for Earth pressure, we want to consider the pressure at the
agricultural hub of the world Central Valley California which is actually higher in latitude than
Mesopotamia of the Fertile Crescent but is where Turkey is of the Fertile Crescent. It is at +40
degrees. We use equations 5.7 and 5.8!
Equation 5.7. Winter: !
Equation 5.8. Summer: !
Equation 5.24. !
We have!
Winter: !
Summer: !
!
And we see that in Central California the Sun is about two times more intense in the summer
than in the winter. In July it hits a high on the average of 82 degrees F (27.78 C) and in January
a low of 34 degrees F (1.11 C). A temperature swing over a year of 48 degrees F.!
We now want to look at this region with the Clausius-Clapeyron relationship.!
!
But we will let Fresno, California represent Central Valley, California, which is at 36.7 degrees
North latitude. We have:!
1.013E5N/m
2
9.8689762E 4
1.013E5
= 0.9742
g =
GM
e
R
2
e
P = G
M
atm
⋅ M
e
4πR
4
e
cos(lat + 23.5
∘
)I
0
= I
s
cos(lat − 23.5
∘
)I
0
= I
s
S
0
=
L
0
4π(R
e
m)
2
= 1361
Wat ts
meter
2
cos(63.5) = 0.4462
cos(16.5) = 0.95882
0.95882
0.4462
= 2.148857
ln
(
P
1
P
2
)
=
ΔH
vap
R
(
1
T
1
−
1
T
2
)
of 42 60
Winter: !
Summer: !
!
The average high was for 1991-2020 in July and was 97.7F (36.5C) and the average low in
January was 40.6F (4.8 C) and this was a temperature swing over a year of 57.1F or 31.7C.
Thus if it is 273+30C=303K, and the vapor pressure of water is 1atm at temperature 373K what
is the vapor pressure? !
!
!
!
We have the vapor pressure at 303K=30C is 0.048 atmospheres and at 373K which is100C is
the boiling point of water, the vapor pressure is 1 atmosphere. To get that in inches of mercury,
multiply by 29.921. We have P1=1.436 inches of mercury for that pressure in atmospheres. To
convert that to millimeters of mercury multiply by 25.4. We have P1=52 mm Hg at 30 degrees C
is 86 degrees F. Boiling point in Fahrenheit is 212 degrees. So this is at about 40% the boiling
point of water. Water has a vapor pressure of 20 torr= 20 mm Hg at room temperature 22 deg
C =72 deg F. Thus we see our answer makes sense. We have 52 torr at 30 deg C. We see
vapor pressure rises rapidly with temperature and is asymptotic at 100 deg C.!
Important to the study of climate is the science of wind, and we study it here only after looking
at pressure because it is caused by pressure gradients. That is by changes in pressure from
higher altitudes to lower altitudes over latitude or longitude, or both. We could say winds are so
important to understand with respect to weather and climate because most of the storms are
blown into land from the oceans because most of the clouds are formed by evaporation of the
oceans. Important winds to consider are the Westerly winds and the Trade winds. The
westerlies come from the west towards and go towards the east in the middle latitudes
between 30 and 60 degrees, where agriculture is so successful. They are called trade winds,
which mainly come from northeast in the Northern Hemisphere and from the southeast in the
Southern Hemisphere, because they have been used for sailing across the world’s oceans for
centuries allowing colonial expansion into the Americas, and trade routes across the Atlantic
and Pacific Oceans.!
The pressure gradient force on a pressure surface due to Earth’s gravity is for its x and y
components:!
Equation 5.25. !
Equation 5.26. !
cos(60.2) = 0.49697
cos(13.2) = 0.0.97358
0.97358
0.49687
= 1.96
ln
(
P
1
1.0
)
= −
40,700
8.3145
(
1
303
−
1
373
)
ln
P
1
1.0
= − 3.0318
P
1
= 0.048at m
G
x
= − g
∂Φ
∂x
G
y
= − g
∂Φ
∂y
of 43 60
Which can be approximated by!
Equation 5.27. !
Equation 5.28. !
is called the Geopotential height. The flow of air, wind, is from higher heights to lower
heights. Thus if the higher pressure is at 5100 meters and the lower pressure is at 5000 meters
and the drop occurs over 100 km from east to west, then !
!
!
Then!
!
But there is another force acting, the Coriolis force which is not really a force at all, but can be
treated like one. If wind blows from south to north, then over the time it takes to travel that
distance, the earth in its rotation has moved out from under it, and when the wind arrives at
North it will appeared to have landed to the right of its supposed destinations. The Coriolis
force is given by!
Equation 5.29. !
Which has components!
Equation 5.30. !
Equation 5.31. !
If the higher pressure is at 5300 meters at 30 deg N and the lower pressure is at 5000 meters at
50 deg N, and you want the wind at latitude 40 deg N, you must use the component equations
for the so-called geostrophic wind!
Equation 5.32. !
Equation 5.33. !
The rotations of Earth using the sidereal day is !
G
x
= − g
ΔΦ
Δy
G
y
= − g
ΔΦ
Δx
Φ
ΔΦ = 100m
Δx = 100,000m
G
x
= − 9.8
m
s
2
100m
100,000m
= − 0.0098m /s
2
f = 2Ω ⋅ sin(ϕ)
F
x
= f ⋅ v
F
y
= − f ⋅ u
u
g
⋅ f = − g ⋅
ΔΦ
Δy
v
g
⋅ f = g ⋅
ΔΦ
Δx
2π /86164.1 =
of 44 60
!
And one degree of latitude is!
!
!
!
!
18m/s!
This is an easterly wind (blowing at 18m/s from west to east). We multiply by 1.93 to get knots
and find it is 34.74 knots. What kind of a force is this?!
!
The density of air is . If this air hits a square meter we have the force is:!
!
The prevailing winds from 30 to 60 degrees are known as westerlies. Tropical eateries are from
0 to 30 degrees known as Trade Winds.!
Ω = 7.292 × 10
−5
ra d
s
2π R
e
= 2π(6.37E6m)/360 = 1.11E5m
2Ωsin(40
∘
) = 9.3744 × 10
−5
sec
−1
g/f = 9.8/7.292E − 5 = 1.344E5m /s
ΔΦ = − 300m /(50 − 30) = 15m /(de gree)
(1.34E5m /s)(15m /deg)/(1.11E5m /deg) =
F = Aρv
2
ρ = 1.229kg/m
3
F = (1.00m
2
)(1.229kg /m
3
)(18m /s)
2
= 398.2N
of 45 60
6.0 Constant k and Proton Radius To bridge the microcosmos to the macrocosmos we
introduce a constant k.!
Warren Giordano wrote in his paper The Fine Structure Constant And The Gravitational
Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where
‘ ’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational
constant numerically, but neglecting any units.
This follows from what Warren Giordano noticed that
Equation 6.1
We can eliminate and introduce the 6 of six-fold symmetry by introducing Avogadro’s
Number
We make an equation of state for the periodic table:
We can say for any element
Where is the number of protons in the element, so for carbon
Thus by equation 6.1 we have
Equation 6.2 .
h
1 + α
α
10
23
h(1 + α) ⋅ 10
23
= G
10
23
N
A
= 6E 23
ℍ = 1
gra m
atom
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
=
Z ⋅ 6E 23pr oton s
Z ⋅ gr a m s
𝔼 =
Z ⋅ gr a m s
Z ⋅ protons
Z
ℂ
ℂ =
6gra m s
6pr oton s
N
A
=
6(6E 23pr oton s)
6gra m s
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
of 46 60
In the next step to bridge the microcosmos to the macrocosmos we create an intermediary
mass midway between the proton mass and the upper limit of mass for a white dwarf
star to form without collapsing into a blackhole star. It balances with its gravity by radiation
pressure alone, the so-called Chandrasekhar limit (See Appendix 3 for how it is derived):!
Equation 6.3
We make the approximation and define with the geometric mean:
Equation 6.4
Giving
Equation 6.5
Thus by equation 1.8 and equation 1.11 and using , we have
Equation 6.6
We had
We can hone this by reintroducing 0.77 for 3/4
Thus precisely:
We have
We have honing our
m
i
m
p
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
0.77 ≈ 3/4
m
i
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
of 47 60
Thus since we said with our estimate
We have a honed value of
This is our constant for bridging the microcosmos with the macrocosmos. We define the Earth
to be the ground state for the solar system yielding our six-fold basis for Nature
Equation 6.7
Where is the average orbital velocity of the Earth.
We note the ground-state for an electron in a hydrogen atom gives
Where is the potential energy of the electron and is its rest energy.
Thus we have the radius of a proton is given by carbon by evaluating at one second:
But to get that we have to multiply by one second and we need one second in terms of the atom
for a theory of the proton. I find we can do that…
Substitute for to get
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
k =
1
(68.897kg)
2
(6.62607E − 34) ⋅
(1.007299)
6.67408E − 11
⋅ 6.02E 23 = 0.001268291s /m
1
k
= 788.4626
m
s
k v
e
= 6
k v
e
=
29790m /s
788.46m /s
= 6.145748
v
e
α
2
=
U
e
m
e
c
2
U
e
m
e
c
2
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
t
6
=
1
α
2
m
p
h 4π r
2
p
Gc
t
6
=
r
p
α
2
m
p
h 4π
Gc
R
H
/2
r
p
of 48 60
Where is the Van Der Waals radius for a hydrogen atom (See Appendix 2). We have
now introduced the radius of a hydrogen atom . Our formulation of inertia as
proton seconds is a form of impulse. To change that to momentum we have to divide by a
second. This radius of the hydrogen atom is the Van Der Waals radius, which is the closest
distance between two hydrogen atoms noncovalently bound. It is 120 pm. Divide that by
where 1/k is our constant
And we find
We have our equation for the radius of a proton
We only need to multiply it by to have the right units, and we get
Equation 6.8
Then suggest we picked up 9/8 in approximations which is close to one anyway so we write
Equation 6.9
We form constants:
t =
R
H
2α
2
m
p
h 4π
Gc
R
H
R
H
= 1.2E − 10m
ck
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon ds
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
2
p
⋅
R
H
N
A
𝔼
= 1.12secon ds
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
t /ck = 1secon d
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
= 8.26935E − 16m ≈ 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
ℍ
=
1
kg
(
kg
)
m
of 49 60
And we have the Equation:
Equation 6.10
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
Equation 6.11.
If our equation is right and we put it into natural units then the product should be close to
one:
Let us start with the units with which we are working:
And convert these to proton-masses and proton-radii:
Now we find k in these units:
k =
hc
2π
3
G
= 6.93E − 9kg
R
H
N
A
𝔼
= 1.99E − 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
m
p
G =
m
3
kg ⋅ s
2
h = kg ⋅
m
2
s
c = m /s
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅ 1.67262E − 27kg(0.833E − 15m)
3
⋅ s = 193,131, 756
h = 6.62607E − 34kg ⋅
m
2
s
⋅
s
(0.833E − 15)
2
(1.67262E − 27kg)
= 5.71E 23
c =
(299,792, 459m /s)(1sec)
(0.833E − 15m)
= 3.6E 23
R
H
=
1.2E − 10m
0.833E − 15m
= 144,058
of 50 60
Thus we have:
Equation 6.12
!
We will say our theory is holistic and that we are describing the proton radius in terms of the
whole of which it is a part, namely, the radius of a hydrogen atom, more specifically the Van Der
Waals radius, which is determined by hydrogen gas, or (See Appendix 3 for the theory of the
Van Der Waals radius).
k =
hc
2π
3
G
= 6.93E − 9kg
k =
(5.71E 23)(3.6E 23)
2π
3
(193131756)
= 4E18pr oton − m a sses
r
p
m
p
= k
R
H
N
A
𝔼
r
p
m
p
=
(4E18)(144058)
(6E 23)
=
5.76E 23
(6E 23)
= 0.96 ≈ 1
H
2
of 51 60
7.0 The Proton Charge
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
Equation 7.1
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
x
ℝ
3
(x, y, z)
ℝ
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
−x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
− ct
2
)
i
··
x ∝ (ct
1
− ct
2
)
i
x
ℝ
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
− ct
2
)
i
g ⋅ m =
m
s
2
Fig. 2
of 52 60
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from our equation for k
And we have
Equation 7.2
We get
Equation 7.3
g =
1
s
2
=
1
t
2
1
s
2
⋅
m
s
⋅ s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
⋅
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
⋅
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
h 4π r
2
p
Gc
=
(6.626E − 34)(4π)(0.833E − 15)
2
(6.674E − 11)(299,792, 459)
= 5.37E − 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E − 31) = 9.1435E − 19C
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71proton s ≈ 6proton s
of 53 60
Appendix 1 Kinetic Energies of Moon and Earth
We suggested the second was a natural unit, and that, if it was, should be in the orbital
mechanics of the earth because the second comes from the calendar, which is based on the
orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the
rotation of the Earth. We found that it was as the following:
Let’s show that…
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion we have:
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion.
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 54 60
Appendix 2 Van Der Waals Radius
Johannes Diderik van der Waals (1873) described more than just Ideal gases, which are gases
that behave according to kinetic-molecular theory, he described real gases which don’t. His
equation then, The Van der Waals equation, is a modification of the Ideal Gas Law which is:
Which is quite obvious. If you increase the temperature T, then the volume of the gas is going to
increase, and if it doesn’t then the pressure will, which is inversely proportional to volume.
However for a Real Gas, he assumed the particles are hard spheres, cannot be compressed
beyond a limit, and at close proximity to one another they interact and have a volume around
them that excludes one another, that is they have walls. He said
That is the volume of the real gas ( ) is equal to the volume of the ideal gas ( ) minus a
correction factor b. The volume of the particles is the number of particles ( ) times the volume
of one particle:
Thus there exists a sphere of radius 2r formed by two particles in contact where no other
particles can enter. It gives the correction factor
PV = n RT
V
R
= V
I
− b
V
R
V
I
n
n
4
3
π r
3
of 55 60
And the volume correction for n particles is
This is the volume correction to the Ideal Gas Law. The pressure correction says real gases
exhibit less pressure because their particles interact which is a net pulling by the bulk of
particles away from the container walls.
The reduction in pressure is proportional to by a factor a. We have for reduction of pressure
that
We substitute this into the Ideal Gas Law:
This can be written as a cubic
Which allows one to compute the critical conditions of liquefaction and to derive an expression
of the principle corresponding states. In the cubic form we have as the solution three volumes
which can be used for computing the volume at and below critical temperatures.
Thus the Van Der Waals radius is estimated
For hydrogen experimentally. Therefore with
We have described the derivation of radius of a hydrogen atom from the Van Der Waals
equations that we use to get
b = (4)
4
3
π r
3
nb = 4n ×
4
3
π r
3
n
2
v
2
P
I
= P
R
+ a
n
2
V
2
(
P + a
n
2
V
2
)
(
V − nb
)
= n RT
V
3
−
(
b +
RT
P
)
V
2
+
a
p
V −
a b
P
= 0
(
π +
3
φ
2
)
(3φ − 1) = 8τ : π =
P
P
c
, φ =
V
V
c
, τ =
T
T
c
4
3
π r
3
w
=
b
N
A
r
3
w
=
3
4π
⋅
b
N
A
b = 26.61
cm
3
m ol
N
A
= 6.02E 23
r
w
= 1.0967E − 8cm = 1.0967E − 10m
of 56 60
Where the Van Der Waals equations are
2
6
R
H
h
πα
2
m
2
p
GN
A
= 1secon d
(
P + a
n
2
V
2
)
(
V − nb
)
= n RT
V
3
−
(
b +
RT
P
)
V
2
+
a
p
V −
a b
P
= 0
4
3
π r
3
w
=
b
N
A
of 57 60
Appendix 3 Chandrasekhar Limit
The pressure of the outer shell of star balances with the outward pressure in the core of
the star (thermal pressure). Pressure is force per unit surface area thus…
is the mass of the core pulling in the mass of the shell and is the radius of the
core. The surface area of the star is that of a sphere, . We have
The thermal pressure countering the gravity is given by the ideal gas law PV=nRT (pressure
times volume of a gas such as hydrogen , which is all protons , is proportional to temperature.
The number of protons in the core is . We have
Where is the Boltzmann constant ( ). Since we must have
if the star is not to implode or explode
And we have the estimate for the temperature of the core of a star.
Fusion would not occur at the low temperature of a star like the Sun in that there would not be
enough energy for collisions, unless the potential Coulomb barrier can be overcome by quantum
mechanical tunneling. The collisions are given by the kinetic energy of the particles
. We have
The velocity v yields the minimum distance between protons as the De Broglie wavelength
P
gravity
P =
F
A
F = m a = PA
P =
m a
A
m a = G
M
shell
M
core
r
2
core
M
core
M
shell
r
core
A = 4π r
2
core
P
gravity
= G
M
shell
M
core
4π r
4
core
m
p
N
p
M
core
m
p
P
thermal
=
M
core
m
p
⋅
1
4
3
π r
3
core
k
B
T
core
k
B
1.380649E − 23J ⋅ K
−1
P
gravity
= P
thermal
k
B
T
core
=
1
3
GM
shell
m
p
r
core
1
2
m
2
p
v
e
2
4πϵ
0
r
min
=
1
2
m
p
v
2
of 58 60
Since the velocity is the root mean square velocity of the protons…
We have the temperature of the star is
This is another estimate. Since the mass of a star is its volume times its density
But for a star density varies with radius
If we take the derivative of both sides of the equation we have one of the equations of stellar
structure:
1.
The so-called conservation of mass equation. The force on the shell of the star is given by the
mass of the shell
Again for there to be balance gravitation pressure equals thermal pressure:
2.
Another equation of the equations of stellar structure. The so-called equation of hydrostatic
equilibrium. This can be written
If the star is an ideal gas the density of the star varies as where for a
monatomic gas and then
λ =
h
m
2
p
v
v
rms
=
3k
B
T
mp
T
min
=
(
e
2
4πϵ
0
)
2
m
p
3π
2
h
2
k
B
m =
4
3
π r
2
ρ
4π
∫
r
0
r
2
ρ(r)dr
dm(r)
dr
= 4π r
2
ρ(r)
F
g
= − G
m(r)4π r
2
ρ(r)
r
2
dr
dρ(r)
dr
= − G
m(r)ρ(r)
r
2
d
dr
(
r
2
ρ(r)
dP(r)
dr
)
= − 4π Gr
2
ρr
PV
γ
= constant
γ = 5/3
of 59 60
In stellar dynamics we write
So that
The abundance of hydrogen and helium in the universe are approximately 75% and 24%,
respectively. Thus for every 4He2+ there are 12H+ and 2+12 free electrons. We have
Ionized hydrogen and helium have and for the Sun because of high metal
content. Finally stars can be approximated as blackbody radiators (purely radiate) and as such
pressure is given in terms of temperature (Temperature is proportional to radiation energy):
There are three kinds of pressures that can be generated by a star: gas pressure, radiation
pressure, or degeneracy pressure.
A type of star that is stable, that is prevented from collapse by degeneracy pressure, is a so-called
white dwarf star. They are the remnant of giant stars that have depleted the their fusion fuel and
thereby collapsed under gravity but are kept from collapsing into black holes by thermal
pressure due to motion of the particles alone. Interestingly, they still shine almost as bright as a
star on the main sequence even though they are not doing fusion. It was the Indian physicist
Chandrasekhar who found the limit in mass for which a white dwarf will not have its gravity
overcome the degeneracy pressure and collapse. The non-relativistic equation is:
There are many resources available that derive this and you can find it in any textbook on
astrophysics in the chapters dealing with stellar physics, and I will leave the treatment of the
derivation to those works.
P ∝
1
V
γ
∝ ρ
5/3
N
V
=
ρ
μm
p
P
gas
=
ρ
μm
p
k
B
T
4 + 12
1 + 12 + 14
= 0.59
μ = 0.59
μ = 0.62
P
rad
=
4
3
⋅
σ
c
T
4
M ≤ 0.77
c
3
h
3
G
3
N
m
4
p
= 1.41 ⊙
of 60 60
The Author!
